Question

a company uses two types of raw material, 1 and 2 for its product. If it uses x units of 1 and y units of 2, it can produce U units of finished items, where U(x,y)=8xy+32x+40y-〖4x〗^2-〖6y〗^2. Each units of 1 costs RM 10 and each units of 2 costs RM 4. Each unit of product can be sold for RM 40. How can the company maximize its profits?

Answer 1

$$U(x,y)=8xy+32x+40y-4x^2-6y^2$$

$$R(x, y)=40(8xy+32x+40y-4x^2-6y^2)$$

$$C(x, y)=10x+4y$$

$$P(x,y)=R(x, y)-C(x, y)$$

$$P(x, y)=40(8xy+32x+40y-4x^2-6y^2)-10x-4y$$

$$P_x=320y+1280-320x-10$$

$$P_y=320x+1600-480y-4$$

Find the critical point(s)

$$P_x=0, P_y=0$$

$$320y+1280-320x-10=0$$ $$320x+1600-480y-4=0$$

$$x=y+{127\over 32}$$ $$160y=2866$$

$$x={3501\over 160}\approx22$$

$$y={2866\over 160}\approx18$$

$$P_{xx}=-320\lt 0$$ $$P_{yy}=-480$$

$$P_{xy}=P_{yx}=320$$

$$D=\begin{vmatrix} -320 & 320 \\ 320 & -480 \end{vmatrix}=51200\gt 0$$

$D\gt 0, P_{xx}\lt 0$. Hence, $P({3501\over 160}, {2866\over 160})$ is a local maximum. Since P(x, y) has the only extrema, then $P({3501\over 160}, {2866\over 160})$ is the absolute maximum.

The company may use 22 units of 1 and 18 units of 2 to maximize its profit.

$$P(22, 18)=RM \ 28188.$$